# How do you solve log_2 (4x)=5?

Jul 11, 2016

8

#### Explanation:

in a variety of ways

for example

${\log}_{2} \left(4 x\right) = 5$

$\implies {\log}_{2} 4 + {\log}_{2} x = 5$
$\implies 2 + {\log}_{2} x = 5$
$\implies {\log}_{2} x = 3$
$\implies x = {2}^{3} = 8$

OR

${\log}_{2} \left(4 x\right) = 5$
${2}^{{\log}_{2} \left(4 x\right)} = {2}^{5}$
$4 x = 32$
$x = 8$

OR

simplest, maybe, taken straight from the idea that a logarithm is just an index

${\log}_{2} \left(4 x\right) = 5$
$\implies 4 x = {2}^{5}$
$\implies x = {2}^{5} \cdot {2}^{- 2} = {2}^{3} = 8$